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TASSpecialist MathematicsSyllabus dot point

How do we describe position, direction, lines and planes in three-dimensional space using vectors?

Use 3D vectors with dot and cross products to find angles, projections, lines and planes.

Three-dimensional vectors: magnitude, dot and cross products, angles, scalar projection, and equations of lines and planes for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.78 min answer

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What this dot point is asking

In three dimensions a vector has three components, a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}, often written as a column (a1,a2,a3)(a_1, a_2, a_3). The basic operations carry over from two dimensions, but two new ideas become essential: the cross product, and the description of lines and planes.

Magnitude and unit vectors

The magnitude (length) is

∣a∣=a12+a22+a32. |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

A unit vector in the direction of a\mathbf{a} is a^=a∣a∣\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}. Unit vectors are useful whenever you only care about direction, for example when writing the direction of a line.

The dot (scalar) product

aβ‹…b=a1b1+a2b2+a3b3=∣aβˆ£β€‰βˆ£b∣cos⁑θ, \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta,

where ΞΈ\theta is the angle between the vectors. Rearranging gives the angle:

cos⁑θ=aβ‹…b∣aβˆ£β€‰βˆ£b∣. \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}.

Two non-zero vectors are perpendicular exactly when aβ‹…b=0\mathbf{a}\cdot\mathbf{b} = 0. The scalar projection of a\mathbf{a} onto b\mathbf{b} is aβ‹…b∣b∣\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}, and the vector projection is aβ‹…b∣b∣2 b\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}.

The cross (vector) product

The cross product produces a vector perpendicular to both inputs:

aΓ—b=∣ijka1a2a3b1b2b3∣=(a2b3βˆ’a3b2)iβˆ’(a1b3βˆ’a3b1)j+(a1b2βˆ’a2b1)k. \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}.

Its magnitude is ∣aΓ—b∣=∣aβˆ£β€‰βˆ£b∣sin⁑θ|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta, which equals the area of the parallelogram spanned by a\mathbf{a} and b\mathbf{b}. Note that aΓ—b=βˆ’(bΓ—a)\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a}), so order matters.

Lines and planes

The vector equation of a line through point r0\mathbf{r}_0 with direction d\mathbf{d} is

r=r0+td,t∈R. \mathbf{r} = \mathbf{r}_0 + t\mathbf{d}, \qquad t \in \mathbb{R}.

The equation of a plane with normal n=(a,b,c)\mathbf{n} = (a,b,c) through point r0\mathbf{r}_0 is nβ‹…(rβˆ’r0)=0\mathbf{n}\cdot(\mathbf{r} - \mathbf{r}_0) = 0, which expands to the Cartesian form ax+by+cz=dax + by + cz = d where d=nβ‹…r0d = \mathbf{n}\cdot\mathbf{r}_0. To find a normal to a plane through three points, take the cross product of two vectors lying in the plane.

Putting it together

Many exam questions chain these tools: find a direction with subtraction, a normal with a cross product, an angle with a dot product, then assemble a line or plane equation. Lay your working out in clear named steps so each result is easy to reuse later in the question.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 TASC6 marksThe point P(1, -2, 3) lies on the plane Pi: 3x + 4y - 2z = d. (a) Determine d. (b) Write the line L through P parallel to direction (2, mu, -1) in parametric form. (c) If L is embedded in Pi, determine mu.
Show worked answer β†’

(a) Substitute P into the plane equation: 3(1) + 4(-2) - 2(3) = 3 - 8 - 6 = -11. So d = -11. (2 marks)

(b) A line through P with direction (2, mu, -1) is (x, y, z) = (1, -2, 3) + t(2, mu, -1), i.e. x = 1 + 2t, y = -2 + mu t, z = 3 - t. (2 marks)

(c) If L lies in the plane, its direction must be perpendicular to the plane's normal (3, 4, -2). Set the dot product to zero: 3(2) + 4(mu) + (-2)(-1) = 6 + 4mu + 2 = 0, so 4mu = -8 and mu = -2. (Because P is already on the plane, this condition is sufficient.) (2 marks)

2024 TASC5 marksPlanes Pi1: x + 2y - 3z = 4 and Pi2: 2x + 4y + cz = d. (a) Show P(3,2,1) lies on Pi1 and Q(c, d/4, -2) lies on Pi2. (b) For what value of c are the planes parallel? (c) With this c and d = 8, prove Q also lies on Pi1.
Show worked answer β†’

(a) For P on Pi1: 3 + 2(2) - 3(1) = 3 + 4 - 3 = 4, true. For Q on Pi2: 2c + 4(d/4) + c(-2) = 2c + d - 2c = d, true. (2 marks)

(b) Two planes are parallel when their normals are scalar multiples. Normal of Pi1 is (1, 2, -3); normal of Pi2 is (2, 4, c). Matching (2, 4, c) = 2(1, 2, -3) gives c = -6. (1 mark)

(c) With c = -6 and d = 8, Q = (-6, 2, -2). Substitute into Pi1: (-6) + 2(2) - 3(-2) = -6 + 4 + 6 = 4, which equals the right side, so Q lies on Pi1. (2 marks)

2021 TASC6 marksQ is the plane 4x - 3y + 6z = 36. (a) Show A(3, 2, 5) lies on Q. (b) Determine the equation of the line joining A to B(1, 2, -3). (c) Find the equation of the plane parallel to Q that contains B.
Show worked answer β†’

(a) Substitute A: 4(3) - 3(2) + 6(5) = 12 - 6 + 30 = 36, which matches, so A lies on Q. (1 mark)

(b) Direction AB = B - A = (1 - 3, 2 - 2, -3 - 5) = (-2, 0, -8), or simplified (1, 0, 4). The line is (x, y, z) = (3, 2, 5) + t(1, 0, 4), i.e. x = 3 + t, y = 2, z = 5 + 4t. (3 marks)

(c) A parallel plane has the same normal (4, -3, 6), so it has the form 4x - 3y + 6z = k. Substitute B(1, 2, -3): 4 - 6 - 18 = -20. The plane is 4x - 3y + 6z = -20. (2 marks)